Physics 161: Black Holes: Lecture 12: 1 Feb 2010
Professor: Kim Griest
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12.1
Orbits in the Schwarzschild metric continued
Effective Potential for Schwarzschild Orbits
Let’s apply the same effective potential technique to the full r geodesic equation we derived earlier.
m
dr
dτ
2
=
l2
E2
rS 2
mc
+
,
−
1
−
mc2
r
mr2
where rS = 2GM/c2 . Here the effective potential is
rS l2
Vef f = 1 −
mc2 +
,
r
mr2
and keep in mind that we are using the proper time τ , not t as the time variable, and l is the angular
momentum and E is the total energy. We can spot the different types of orbits as we did for the Newtonian
effective potential. Now we see there are five different types of orbits.
Fig: Effective potential for Schwarzschild Geodesics
1. At the minimum we have a bound circular orbit. The radius is found from dVef f /dr = 0 as before.
2. We also have the bound orbits with turning points r1 and r2 like in the Newtonian case, but here
these turn out to not be elliptical! The orbits don’t close. If we do the perhelion advance of Mercury
we will show this. The time to between r1 and r2 and back is more than the time to go around
φ = 3600, so the perhelion advances.
3. Next, as for Newtonian hyperbolic orbits, we have orbits that start at r = ∞, come in, reach a
distance of closest approach, and go out again. Close examination shows these are close to, but not
exactly the same as, the Newtonian case.
4. Now look at maximum point on top of the hill. We find this point also when we set dVef f /dr = 0,
since the slope is zero here. If you set a particle there and carefully balanced it, it would stay there.
Thus this is another circular orbit, but it is unstable. A tiny perturbation outward and the particle
will escape to infinity. A tiny perturbation inward and it will fall in the hole.
5. Finally, there is another new type of orbit not found in the Newtonian case, the capture orbit. If
the energy is high enough, the particle goes over the high point and then plunges down into the
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hole, never to return. In the Newtonian case a particle can never be captured in this way due to
gravity alone, even though this is how we think when, for example, a comet hits the Sun. Hitting
the Sun’s surface invokes non-gravitational forces; a point mass object could never capture anything
in Newtonian mechanics.
Let’s find the radii of the circular orbits. We can write
Vef f = m +
l2
rS m rS l2
−
,
−
mr2
r
mr3
and differentiate and set to zero.
rS m 3rS l2
2l2
dVef f
= 0.
=− 3 + 2 +
dr
mr
r
mr4
We can simplify this to a quadratic equation in r by multiplying through by mr4 :
rS m2 c2 r2 − 2l2 r + 3rs l2 = 0,
which has two solutions:
l2
r± =
rS m2 c2
1±
r
3r2 m2 c2
1− S 2
l
!
,
where I’ve put the c’s back in for fun and reference.
Note that if the quantity in the square root is negative we don’t have a real solution. Thus there is
no circular orbit in that case. This happens when the angular momentum is too small, i.e. there are
only capture orbits. If the quantity in the square root in positive, then we have two solutions, that is,
two circular orbits at different radii, as we saw in the plot. Thus we have two circular orbit solutions if
l2 > 3rS2 m2 c2 , and none if not.
When there are two orbits we can tell if the orbits are stable or unstable by taking the 2nd derivative.
If d2 Vef f /dr2 > 0, the the curvature of the effective potential is positive and it is a minimum. This
means the orbit is stable. On the other hand if if d2 Vef f /dr2 < 0, it means there is a maximum in the
effective potential at that point, and the orbit is unstable. We can do the math, but we saw before from
the picture that the smaller solution (the one with the minus square root) was unstable, and the larger
solution with the plus square is a stable orbit.
From the above discussion we see that the smallest possible stable circular orbit will happen when
the square root term vanishes. This happens when 1 = 3rS m2 c2 /l2 = 0, or l2 = 3rS2 m2 c2 . The value of
the radius for this value of the angular momentum can be found just by plugging this value of l into the
formula for r± , and we find that the minimum stable circular radius is
rmin =
l2
= 3rS ,
rS m2 c2
that is just exactly three times the Schwarzschild radius. Actually we should check that this orbit is
stable, since for this value of l2 , the stable and unstable orbits are the same. Plugging in rmin , and the
value of l2 above into
6l2
2rS m 12rS l2
d2 Vef f
=
−
−
,
dr2
mr4
r3
mr5
2
we find that d2 Vef f /dr = 0 at this combined minimum. Thus this orbit is just really neither stable nor
unstable, but neutral. It is called, however, the minimum or last stable orbit, since an orbit just a tiny
bit larger is in fact stable.
This last stable orbit value is a very important result. When things fall into black holes they have
trouble getting in because of angular momentum. The tend to get ripped apart and form what are called
accretion disks. The material in the disk gradually spirals inward. However, when the radius reaches
3rS , there is no longer a stable circular orbit, so the material all just flows into the hole. Thus we expect
that when we look at real black holes in space we will see material down to 3rS but not any closer. When
we detect X-rays from black holes, we hope we are looking at material radiating from the distance 3rS .
And when we find periodic signals from black hole candidates, we assume this is the radius at which the
material is orbiting.
12.2
Producing energy using black holes
Now let me tell you why black holes are the most efficient energy generation devices we have yet concieved.
Much better than nuclear fission or even nuclear fusion. Of course, there is the small problem of not
having a black hole near by and available for use.
First consider throwing some junk into a black hole from far away. We learned before that the total
energy of the stuff far from the black hole is just E = mc2 , where m is the mass of the stuff. Suppose
this stuff doesn’t go straight into the hole, but is slowed down in a series of orbits, finally settling into the
last stable orbit we just discussed. What is the energy of the stuff now? We can use the radial geodesic
equation to find out. On a circular orbit dr/dτ = 0, so we have
l2
rS E2
m+
.
= 1−
m
r
mr2
Next we plug in l2 = 3rS2 m2 , and r = 3rS ,
E2
1
3r2 m2
1
1
8
= (1 − )(m + S 2 ) = (1 − )(1 + )m = m.
m
3
m9rS
3
3
9
Thus
r
8 2
mc .
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Somehow during the transition from far away to the last stable orbit the energy has decreased. That
energy must have been radiated away, and
q therefore be available for other use. The amount of energy
E=
radiated away is (mc2 − E)/mc2 = 1 − 89 = 5.7%.
Let’s compare this with other forms of energy generation. When burned, a gallon of gasoline produces
1.32 × 108 Joules of energy and has a mass of 2.8 kg. Thus the fractional mass energy lost when a gallon
of gasoline is burned is ∆E/E = (1.32 × 108 )/((2.8)(3 × 108 )2 ) = 5.2 × 10−10 . This is thus also the
fraction of the rest mass of the gasoline that is turned into useful energy, a far cry from the 5.7% we get
from the black hole.
Next, consider nuclear energy, say the complete fission of 1 kg of pure Uranium 235. This produces
8.28 × 1013 Joules, so ∆E/E = (8.28 × 1013)/((1)(3 × 108 )2 ) = 0.0092%, still below the black hole. Finally
one usually thinks of the main source of energy in the Universe as nuclear fusion of Hydrogen into Helium
(as in the Sun). Actually in large stars, the Helium is then burned to Carbon, Oxygen, etc. all the way
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up to Iron, which is the most stable element. This whole chain of nuclear fusion releases about 0.9% of
the rest mass into usable energy, still a factor of six below what you get from throwing your garbage into
a black hole. Black holes are the best devices we know of for producing energy. In fact, using a rotating
black hole we can turn even more of the rest mass of stuff into useful energy. The energy generation
method discussed here is thought to be the power source of quasars, the most luminous and energetic
steady objects in the Universe.
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